Q. 13 4.5( 141 Votes )

D and E are points on the sides CA and CB respectively of a triangle ABC right angled at C. Prove that AE2 + BD2 = AB2 + DE2

Answer :

To Prove:  AE2 + BD2 = AB2 + DE2

Given: D and E are midpoints of AD and CB and ABC is right angled at C
Applying Pythagoras theorem in ΔACE, we obtain

Pythagoras theorem: It states that the square of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the other two sides.

AC2 + CE2 = AE2           ..........eqn(i)

Applying Pythagoras theorem in triangle BCD, we get

BC2 + CD2 = BD2            .........eqn(ii)

Adding equations  (i) and (ii), we get

AC2 + CE2 + BC2 + CD2 = AE2 + BD2      .................eqn (iii)

Applying Pythagoras theorem in triangle CDE, we get

DE2 = CD2 + CE2

Applying Pythagoras in triangle ABC, we get

AB2 = AC2 + CB2

Putting these values in eqn(iii), we get

DE2 + AB2 = AE2 + BD2

Hence, Proved.

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